Let $R$ be the region enclosed by the line $y=x$, the curve $y=e^{x-1}$, and the line $x=3$. $y$ $x$ ${y=e^{x-1}}$ ${y=x}$ $y=-1}$ $ 3$ $(1,1)$ $ R$ A solid is generated by rotating $R$ about the line $y=-1$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_{-1}^3 \left( e^{2x-2}-x^2-1 \right)dx$ (Choice B) B $\pi \int_{-1}^3 \left( e^{2x-2}+2e^{x-1}-x^2-2x \right)dx$ (Choice C) C $\pi \int_1^3 \left( e^{2x-2}-x^2-1 \right)dx$ (Choice D) D $\pi \int_1^3 \left( e^{2x-2}+2e^{x-1}-x^2-2x \right)dx$
Answer: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=e^{x-1}}$ ${y=x}$ $y=-1}$ $ 3$ $(1,1)$ Let the thickness of each slice be $dx$, let the radius of the washer, as a function of $x$, be $r_1(x)$, and let the radius of the hole, as a function of $x$, be $r_2(x)$. Then, the volume of each slice is $\pi[(r_1(x))^2-(r_2(x))^2]\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(x))^2-(r_2(x))^2]\,dx$ This is called the washer method. What we now need is to figure out the expressions of $r_1(x)$ and $r_2(x)$, and the interval of integration. $r_1(x)$ is equal to the distance between the curve $y=e^{x-1}$ and the line $y=-1$. So, ${r_1(x)=e^{x-1}+1}$. $r_2(x)$ is equal to the distance between the line $y=x$ and the line $y=-1$. So, ${r_2(x)=x+1}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(x)})^2-({r_2(x)})^2] \\\\ &= \pi\left[ \left( {e^{x-1}+1} \right)^2-\left[ {x+1} \right]^2 \right] \\\\ &=\pi\left[ \left( e^{2x-2}+2e^{x-1}+1 \right)-\left(x^2+2x+1\right) \right] \\\\ &=\pi\left( e^{2x-2}+2e^{x-1}-x^2-2x \right) \end{aligned}$ The leftmost endpoint of $R$ is at $x=1$ and the rightmost endpoint is at $x=3$. So the interval of integration is $[1,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^3 \pi\left( e^{2x-2}+2e^{x-1}-x^2-2x \right)dx \\\\ &=\pi \int_1^3 \left( e^{2x-2}+2e^{x-1}-x^2-2x \right)dx \end{aligned}$